Difference between revisions of "2024 AMC 12A Problems/Problem 16"
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==Solution 1A (Trivial/Easy solve)== | ==Solution 1A (Trivial/Easy solve)== | ||
| − | We have <math>\binom{12}{4,4,4}</math> ways to handle the red/white/ | + | We have <math>\binom{12}{4,4,4}</math> ways to handle the red/white/blue balls distribution on the denominator. |
Now we simply <math>\binom{6}{1}</math> <math>\binom{5}{2}</math> <math>3!</math> for the numerator in order to handle the black balls and distinguishable persons. | Now we simply <math>\binom{6}{1}</math> <math>\binom{5}{2}</math> <math>3!</math> for the numerator in order to handle the black balls and distinguishable persons. | ||
The solution is therefore <math>\frac {6 \cdot 6 \cdot 10}{70 \cdot 45 \cdot 11} = \frac {4}{385}</math> or <math>4+385=\boxed{\textbf{(C) }389}.</math> | The solution is therefore <math>\frac {6 \cdot 6 \cdot 10}{70 \cdot 45 \cdot 11} = \frac {4}{385}</math> or <math>4+385=\boxed{\textbf{(C) }389}.</math> | ||
Revision as of 15:45, 17 November 2024
Contents
Problem
A set of 
tokens — 
red, 
white, 
blue, and 
black — is to be distributed at random to game players, 
tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as 
, where 
and 
are relatively prime positive integers. What is 
?
Solution 1A (Trivial/Easy solve)
We have 
ways to handle the red/white/blue balls distribution on the denominator.
Now we simply 
for the numerator in order to handle the black balls and distinguishable persons.
The solution is therefore 
or 
Remarks - Notice we let balls and persons be indistinguishable to increase ease of calculations
~polya_mouse
Solution 1B (12fact bash)
We have 
total possible arrangements of distinct tokens. If we imagine the first tokens of our arrangement go to the first player, the next go to the second, and the final go to the third, then we can view this problem as counting the number of valid arrangements.
Firstly, the tokens are not all distinct, so we multiply by , 
, 
, and 
to account for the fact that the red, white, blue, and black tokens, respectively can switch around from where they are.
Letting 
denote red, 
denote white, 
denote blue, and 
denote black, then our arrangement must be something like 
. The three players are arbitrary, so we multiply by ; then, the player who gets the reds has 
possible arrangements, the player who gets the whites has 
possibilities, and the player who gets the blacks has 
possibilities. Our total on top is thus 
, and the denominator is . Firstly, we have the in the numerator cancel out part of the denominator; we thus have the following:
![\[\dfrac{3\cdot2\cdot2\cdot3\cdot2\cdot4\cdot6\cdot4}{12\cdot11\cdot10\cdot9\cdot8\cdot7}=\dfrac{2^83^3}{2^63^35\cdot7\cdot11}=\dfrac4{385}.\]](http://latex.artofproblemsolving.com/4/6/2/4626567e9829f4d9c20a59621b51fa34eec8c553.png)
Our answer is
~Technodoggo
Solution 2
Assume all of them are distinct even though some have the same color,
Total possibility = 
(choosing 4 random token for each person)
Next, assume that all the token are already in 3 different groups (Note: 3! Ways to do so since 3 people)
We then distribute the 6 distinct black token into these 3 different groups (So 1,2,3 token for each group)
There are a total of 
ways in doing so
Thus the answer is 
So the answer is 
~lptoggled
Solution 3
We first assume there are designated red, white, and blue token players that will receive all of their respective one.
Consider each non-black token: The probability of the red player getting the first red token is 
, because each player has 4 empty token "slots" for a total of 12. It follows that the probability of the player receiving all 3 red tokens is 
, the white token player is 
, and the blue token player is 
.
The combined probability is 
.
Finally, we multiply the probability by 
to remove our initial assumption to get 
.
The requested sum is 
.
~SilverRush
Solution 4
Process start, first player get tokens. The probability that he gets 3 red tokens and 1 black token is 
There is 
possible arrangements (RRRB,RRBR,RBRR,BRRR) and 3 possibilities who is the first, so the probability that some player gets 3 red tokens and 1 black token is 
After that second player get tokens. The probability that he gets 2 white tokens and 2 black tokens is 
There is 
possible arrangements (WWBB,WBWB, WBBW, BBWW, BWBW, BWWB) and 2 possibilities who is the second, so the probability that some player gets 2 white tokens and 2 black tokens is 
The third player gets last tokens - 1 blue and 3 black tokens.
The desired probability is 
.
To check the result suppose that first and some) player gets 1 blue and 3 black tokens. The probability is 
The probability that second (and some) player gets 3 red tokens and 1 black token is 
The desired probability is 
vladimir.shelomovskii@gmail.com, vvsss
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=ynMtiJuLCNI
See also
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
